# definite integral!

• September 21st 2012, 03:22 AM
lawochekel
definite integral!
$\int_{-\pi}^{\pi} \cos^{2}ny dy$

$\left[\frac{n\cos^{3}ny}{3n}\sin{ny}\right]_{-\pi}^{\pi}$

$\left[\frac{n\cos^{3}n\pi}{3n}\sin{n\pi}+\frac{n\cos^{3} n\pi}{3n}\sin{n\pi}\right]$

$2\left[\frac{n\cos^{3}n\pi}{3n}\sin{n\pi}\right]$

$2\left[0\right]$

pls is 0 equivalent to $2{\pi}$ or $\pi$

thanks!
• September 21st 2012, 03:40 AM
Plato
Re: definite integral!
Quote:

Originally Posted by lawochekel
$\int_{-\pi}^{\pi} \cos^{2}ny dy$

$\left[\frac{n\cos^{3}ny}{3n}\sin{ny}\right]_{-\pi}^{\pi}$

You have do this incorrectly. See here.
• September 21st 2012, 08:32 AM
Soroban
Re: definite integral!
Hello, lawochekel!

You integrated incorrectly.

We need the identity: . $\cos^2\theta \:=\:\tfrac{1}{2}\left(1 + \cos2\theta\right)$

Quote:

$\int_{\text{-}\pi}^{\pi} \cos^2(ny)\,dy$

$\int_{\text{-}\pi}^{\pi} \cos^2(ny)\,dy \;=\;\tfrac{1}{2}\int^{\pi}_{\text{-}\pi}\big[1 + \cos(2ny)\big]\,dy$

. . . . . . . . . . . $=\; \tfrac{1}{2}\big[y + \tfrac{1}{2n}\sin(2ny)\bigg]^{\pi}_{\text{-}\pi}$

. . . . . . . . . . . $=\;\tfrac{1}{2}\left[\pi + \tfrac{1}{2n}\sin(2\pi n)\right] - \tfrac{1}{2}\left[\text{-}\pi + \tfrac{1}{2n}\sin(\text{-}2\pi n)\right]$

. . . . . . . . . . . $=\;\tfrac{1}{2}(\pi + 0) - \tfrac{1}{2}(\text{-}\pi + 0)$

. . . . . . . . . . . $=\; \pi$